Answer to Question #98913 in General Chemistry for Sarah G.

Question

An empty steel container is filled with 0.0500 atm of HF. The system is allowed to reach equilibrium. If Kp = 2.76 for the reaction below, what is the equilibrium partial pressure of H₂?

Answer 1

$2HF = H_2 + F_2$

Suppose the system initially contained 2 mol HF, then:

$K_p = {x^2 \over 2 - 2x}$

$2.76(2 - 2x) = x^2$

$5.52 - 5.52x = x^2$

$x^2 + 5.52x - 5.52 = 0$

$x = 0.86$

Then, the amount of substance substances in the reaction in equilibrium:

$n(HF) = 0.28 mol$

$n(H_2) = n(F_2) = 0.86 mol$

$\chi(H_2) = {0.86 \over 0.86 + 0.86 + 0.28} = 0.43$

Then:

$P_{H_2} = \chi(H_2)*P$

$P_{H_2} = 0.43*0.0500atm = 0.0215atm$

Answer: 0.0215 atm